Question 1076596: A game is played using one die. If the die is rolled and shows 4, the player wins $35. If the die shows any number other than 4, the player wins nothing.
If there is a charge of $7 to play the game, what is the game's expected value?
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! A game is played using one die. If the die is rolled and shows 4, the player wins $35. If the die shows any number other than 4, the player wins nothing.
If there is a charge of $7 to play the game, what is the game's expected value?
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Random "winnings":::...28.....-7
Propabilities::::......1/6....5/6
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E(x) = (1/6)28 + (5/6)7 = (28-35)/6 = -7/6 = -$1.17 (expect to lose $1.17)
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cheers,
Stan H.
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